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= \(\frac{\sqrt{x}-\sqrt{x}+1}{\left(\sqrt{x}-1\right)\sqrt{x}}:\left[\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)-\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}\right]\)
\(=\frac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}:\frac{x-1-x+4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}\)
\(=\frac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}\cdot\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}{3}=\frac{\sqrt{x}-2}{3\sqrt{x}}\)
\(a,\frac{\sqrt{108x^3}}{\sqrt{12x}}=\frac{\sqrt{36.3.x^3}}{\sqrt{3.4.x}}=\frac{6\sqrt{3}.\sqrt{x}^3}{2\sqrt{3}.\sqrt{x}}=3\sqrt{x}^2=3x\)
\(b,\frac{\sqrt{13x^4y^6}}{\sqrt{208x^6y^6}}=\frac{\sqrt{13}.\sqrt{x^4}.\sqrt{y^6}}{\sqrt{16.13}.\sqrt{x^6}.\sqrt{y^6}}=\frac{\sqrt{13}.x^2y^3}{4\sqrt{13}x^3y^3}=\frac{1}{4x}\)
\(c,\frac{x\sqrt{x}+y\sqrt{y}}{\sqrt{x}+\sqrt{y}}-\left(\sqrt{x}+\sqrt{y}\right)^2\)
\(=\frac{\sqrt{x}^3+\sqrt{y}^3}{\sqrt{x}+\sqrt{y}}-\left(x+2\sqrt{xy}+y\right)\)
\(=\frac{\left(\sqrt{x}+\sqrt{y}\right)\left(x-\sqrt{xy}+y\right)}{\sqrt{x}+\sqrt{y}}-x-2\sqrt{xy}-y\)
\(=x-\sqrt{xy}+y-x-2\sqrt{xy}-y=-3\sqrt{xy}\)
\(d,\sqrt{\frac{x-2\sqrt{x}+1}{x+2\sqrt{x}+1}}=\frac{\sqrt{\left(\sqrt{x}-1\right)^2}}{\sqrt{\left(\sqrt{x}+1\right)^2}}=\frac{\sqrt{x}-1}{\sqrt{x}+1}\)
Đk chỗ này là \(\sqrt{x}-1\ge0\Rightarrow\sqrt{x}\ge\sqrt{1}\Rightarrow x\ge1\)nhé
\(e,\frac{x-1}{\sqrt{y}-1}.\sqrt{\frac{\left(y-2\sqrt{y}+1\right)^2}{\left(x-1\right)^4}}=\frac{x-1}{\sqrt{y}-1}.\frac{y-2\sqrt{y}+1}{\left(x-1\right)^2}\)
\(=\frac{\left(x-1\right)\left(\sqrt{y}-1\right)^2}{\left(\sqrt{y}-1\right)\left(x-1\right)^2}=\frac{\sqrt{y}-1}{x-1}\)
a) đk : \(x\ge0\) ; \(x\ne1\)
A=\(\left(\frac{2\sqrt{x}}{\left(x+1\right)\left(\sqrt{x}+1\right)}-\frac{x+1}{\left(\sqrt{x}+1\right)\left(x+1\right)}\right):\left(\frac{\sqrt{x}-1}{\sqrt{x}+1}\right)\)
\(=\left(\frac{-\left(\sqrt{x}-1\right)^2}{\left(x+1\right)\left(\sqrt{x}+1\right)}\right):\left(\frac{\sqrt{x}-1}{\sqrt{x}+1}\right)\) \(=\frac{1-\sqrt{x}}{x+1}\)
b) đk : \(x\ne0;x\ne1\)
B=\(\left(\frac{\left(\sqrt{x}-1\right)^2-\left(\sqrt{x}+1\right)^2}{x-1}\right):\left(\frac{1-x}{2\sqrt{x}}\right)^2\) \(=\left(\frac{-2\sqrt{x}}{x-1}\right):\left(\frac{1-x}{2\sqrt{x}}\right)^2\) \(=\frac{-4x}{\left(x-1\right)^3}\)
a, Với \(x\ge0;x\ne1\)
\(Q=\left(\frac{x-1}{\sqrt{x}-1}-\frac{x\sqrt{x}-1}{x-1}\right):\left(\frac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}+1}+\frac{\sqrt{x}}{\sqrt{x}+1}\right)\)
\(=\left(\sqrt{x}+1-\frac{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{x-1}\right):\left(\frac{x-\sqrt{x}+1}{\sqrt{x}+1}\right)\)
\(=\left(\sqrt{x}+1-\frac{x+\sqrt{x}+1}{\sqrt{x}+1}\right):\left(\frac{x-\sqrt{x}+1}{\sqrt{x}+1}\right)\)
\(=\left(\frac{x+2\sqrt{x}+1-x-\sqrt{x}-1}{\sqrt{x}+1}\right):\left(\frac{x-\sqrt{x}+1}{\sqrt{x}+1}\right)\)
\(=\frac{\sqrt{x}}{x-\sqrt{x}+1}\)
\(B=\left(\frac{\sqrt{x}}{\sqrt{x}-1}-\frac{1}{x-\sqrt{x}}\right)\div\left(\frac{1}{\sqrt{x}+1}+\frac{2}{x-1}\right)\)
\(=\frac{x-1}{\sqrt{x}\left(\sqrt{x}-1\right)}\div\frac{\sqrt{x}-1+2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
\(=\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x-1}\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}\div\frac{\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
\(=\frac{\sqrt{x}+1}{\sqrt{x}}.\left(\sqrt{x}-1\right)\)
\(=\frac{x-1}{\sqrt{x}}\)
a) Ta có: \(M=\left(\frac{\sqrt{x}}{\sqrt{x}-1}-\frac{1}{x-\sqrt{x}}\right):\left(\frac{1}{\sqrt{x}+1}+\frac{2}{\sqrt{x}-1}\right)\)
\(=\left(\frac{x-1}{\sqrt{x}\left(\sqrt{x}-1\right)}\right):\left(\frac{\sqrt{x}-1+2\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right)\)
\(=\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}:\frac{\sqrt{x}-1+2\sqrt{x}+2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
\(=\frac{\sqrt{x}+1}{\sqrt{x}}:\frac{3\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
\(=\frac{\left(\sqrt{x}+1\right)^2\cdot\left(\sqrt{x}-1\right)}{\sqrt{x}\left(3\sqrt{x}+1\right)}\)
b) Để M>0 thì \(\frac{\left(\sqrt{x}+1\right)^2\cdot\left(\sqrt{x}-1\right)}{\sqrt{x}\left(3\sqrt{x}+1\right)}>0\)
mà \(\forall\left\{{}\begin{matrix}x>0\\x\ne1\end{matrix}\right.\), ta luôn có: \(\sqrt{x}\left(3\sqrt{x}+1\right)>0\)
nên \(\left(\sqrt{x}+1\right)^2\cdot\left(\sqrt{x}-1\right)>0\)
mà \(\left(\sqrt{x}+1\right)^2>0\forall0< x\ne1\)
nên \(\sqrt{x}-1>0\)
\(\Leftrightarrow\sqrt{x}>1\)
hay x>1(nhận)
Vậy: để M>0 thì x>1
Đề bài đâu bn ơi
Nếu rút gọn thì mình làm cho
Ta có: \(P=\left(\frac{1}{\sqrt{x}}-\sqrt{x}\right):\left(\frac{1-\sqrt{x}}{\sqrt{x}}+\frac{\sqrt{x}-1}{x+\sqrt{x}}\right)\) ( ĐKXĐ: \(x\ge1\))
\(\Leftrightarrow P=\left(\frac{1-x}{\sqrt{x}}\right):\left(\frac{\left(1-\sqrt{x}\right)\left(1+\sqrt{x}\right)+\sqrt{x}-1}{\sqrt{x}.\left(\sqrt{x}+1\right)}\right)\)
\(\Leftrightarrow P=\frac{1-x}{\sqrt{x}}.\frac{\sqrt{x}.\left(\sqrt{x}+1\right)}{1-x+\sqrt{x}-1}\)
\(\Leftrightarrow P=\left(1-x\right).\frac{\sqrt{x}+1}{\sqrt{x}-x}\)
\(\Leftrightarrow P=\left(1-\sqrt{x}\right)\left(1+\sqrt{x}\right).\frac{\sqrt{x}+1}{\sqrt{x}.\left(1-\sqrt{x}\right)}\)
\(\Leftrightarrow P=\frac{\left(1+\sqrt{x}\right)^2}{\sqrt{x}}\)
\(\Leftrightarrow P=\frac{x+2\sqrt{x}+1}{\sqrt{x}}\)
P=\(\frac{1-x}{\sqrt{x}}:\frac{\left(1-\sqrt{x}\right)\left(\sqrt{x}+1\right)+\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}\left(\sqrt{x}+1\right)}\)
P=\(\frac{1-x}{\sqrt{x}}:\frac{1-x+x-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+1\right)}\)
P=\(\frac{\left(1-\sqrt{x}\right)\left(1+\sqrt{x}\right)}{\sqrt{x}}.\frac{\sqrt{x}\left(\sqrt{x}+1\right)}{1-\sqrt{x}}\)
P=\(\left(\sqrt{x}+1\right)^2\)
P=\(x+2\sqrt{x}+1\)